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3 years ago

Use the figure below.

Mathematics

1 answer:

nevsk [136]3 years ago

4 0

Answer:

∠ACB≅∠CAD, Alternate Interior Angles Theorem

∠BAC≅∠DCA, Alternate Interior Angles Theorem

AC≅AC, Reflexive Property of Congruence

By A-S-A congruence, the △ABC≅△CDA.

Step-by-step explanation:

Given the figure:

We are given that:

Side BC || AD

Side AB || DC

Let us have a look at a few properties first.

Alternate Interior Angle Theorem: It states that when two parallel lines are cut by a line then the alternate angles which are on the interior side are equal to each other.

Reflexive Property of Congruence: It states that a side or angle is always congruent to itself.

Now, let us consider the triangles:

△ABC and △CDA.

Side BC || AD

Therefore,

<em>∠ACB≅∠CAD, Alternate Interior Angles Theorem</em>

Side AB || DC

Therefore,

<em>∠BAC≅∠DCA, Alternate Interior Angles Theorem</em>

<em />

Also, Side AC is common.

AC≅AC, Reflexive Property of Congruence

Therefore, by <em>A-S-A congruence</em>, the △ABC≅△CDA.

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il63 [147K]

The required value after simplification of the s = -16/3. None of these are correct.

Given that,
To simplify $[\frac{x^{2/3}x^{-1/2}}{x\sqrt{x^3}\sqrt[3]{x}}]^2$ and to find the value of s in $x^s$ .

<h3>What is simplification?</h3>

The process in mathematics to operate and interpret the function to make the function simple or more understandable is called simplifying and the process is called simplification.

Simplification,

$=[\frac{x^{2/3}x^{-1/2}}{x\sqrt{x^3}\sqrt[3]{x}}]^2\\= \frac{x^{4/3}x^{-1}}{x^2x^3*{x}^{2/3}}\\= \frac{x^{1/3}}{x^{17/3}}\\=x^{-16/3}$
Comparing with $x^S$
s = -16/3

Thus, the required value of the s = -16/3. None of these are correct.

Learn more about simplification here: brainly.com/question/12501526

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2 years ago

!!I'M BEING TIMED ANSWER QUICKLY!! What is the y-intercept of the quadratic function f(x) = (x – 6)(x – 2)? (0,–6) (0,12) (–8,0)

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The answer is (0,12) as the y-int

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4 years ago

Arithmetic and Geometric Sequences (Context)

V125BC [204]

The formula for compound interest

A = P( 1 + r/n) ^ (nt)

A is the amount in the account at the end

P is the principal balance or the amount initially invested

r is the annual interest rate in decimal form

n is the number of times it is coupounded per year

t is the number of years

A = 1800 ( 1+ .0375/1) ^ (1*6)

A = 1800 ( 1.0375)^6

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Rounding to the nearest cent

A = 2244.92

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1 year ago

Suppose two $20 bills, three $10 bills, one $5 bill, and seven $1 bills are placed in a bag. If you were to pull a bill at rando

Korolek [52]

Answer:

Assuming that the $1 bill was pulled at random, then the expected value of the amount chosen is $\frac{7}{13}$ .

Step-by-step explanation:

From the given question, the bag contains;

$1 bill = 7

$5 bill = 1

$10 bill = 3

$20 bill = 2

Total number of bills in the bag = 13

Pulling a bill at random, the bills would have an expected value as follows:

For $1 bill, the expected value = $\frac{7}{13}$

For $5 bill, expected value = $\frac{1}{13}$

For $10 bill, expected value = $\frac{3}{13}$

For $20 bill, the expected value = $\frac{2}{13}$

Assuming that the $1 bill was pulled at random, then the expected value of the amount chosen is $\frac{7}{13}$ .

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3 years ago

We are given a sequence of five non-zero numbers, where the sum of each term and its neighboring terms is 15 or 25. Find the sum

PSYCHO15rus [73]

We have a sequence that meets the given criteria, and with that information, we want to get the sum of all the terms in the sequence.

We will see that the sum tends to infinity.

So we have 5 terms;

A, B, C, D, E.

We know that the sum of each term and its neighboring terms is 15 or 25.

then:

A + B + C = 15 or 25
B + C + D = 15 or 25
C + D + E = 15 or 25

Now, we want to find the sum of all the terms in the sequence (not only the 5 given).

Then let's assume we write the sum of infinite terms as:

$a_1 + a_2 + a_3 + a_4 + a_5 + a_6 + ...$

Now we group that sum in pairs of 3 consecutive terms, so we get:

$(a_1 + a_2 + a_3) + (a_4 + a_5 + a_6) + ...$

So we will have a sum of infinite of these, and each one of these is equal to 15 or 25 (both positive numbers). So when we sum that infinite times (even if we always have the smaller number, 15) the sum will tend to be infinite.

Then we have:

$(a_1 + a_2 + a_3) + (a_4 + a_5 + a_6) + ... \to \infty$

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3 years ago